s2+22s=0

Solution for s2+22s=0 equation:

s2+22s=0

We add all the numbers together, and all the variables

s^2+22s=0

a = 1; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·1·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*1}=\frac{-44}{2}
=-22
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*1}=\frac{0}{2}
=0
$